PERCENTAGE


 Percentage

INTRODUCTION
The percentage is one of the most important topics in quantitative aptitude.
In combination with ratios, it forms the backbone for various aspects of
Arithmetic. The percentage is of immense importance for calculation-
intensive data interpretation problems. Around four to five problems are
asked from the percentage in CAT and OMETs every year.

Definition Definition

Percentage is the fraction value

out of 100 or you can say that a
fraction with denominator 100 is
called percentage.

yy Percentage is nothing but representation of a fraction.

yy The word percentage is taken from the Latin word “percentum” meaning
per 100 and it is denoted by the symbol %.
Eq. A out of B ?
A
% Value = × 100
B
Or
4 out of 5 ? Rack your Brain
4
% value = × 100 = 80%
5 What is 31.5% of 124?
80
yy Also 80% = ,means 80 out of 100.
100
Hence, % is nothing but a way of representing a fraction.

Concept 1: Conversion of a fraction into percentage

To convert a fraction into percentage, multiply the fraction by 100 and put
% sign.

Key Note

A fraction with denominator 10 is

called as decimal.
Percentage

2.
Example 1: Convert the following fractions into percentage:
1 3
(i) (ii)
5 5
3 5
(iii) (iv) Previous Year’s Question
8 7
Solution:
1 1 In May, John bought the same
(i) → × 100 = 20% amount of rice and the same
5 5
amount of wheat as he had
3 3
(ii) → × 100 = 60% bought in April, but spent 150
5 5
more due to price increase of
3 3 rice and wheat by 20% and 12%,
(iii) → × 100 = 37.5%
8 8 respectively. If John had spent
5 5 500 450 on rice in April, then how
(iv) → × 100 = % = 71.42%
7 7 7 much did he spend on wheat in
May?
Concept 2: Conversion of a percentage into a (1) 590 (2) 580
fraction (3) 560 (4) 570

If we want to convert a percentage value into a

1
fraction, so we have to replace % sign with and reduce the fraction into
100
its simplest form.

Example 2: Express the following percentage as fraction:

1 1
(i) 11 % (ii) 62 %
9 2
(iii) 30% (iv) 40%

Solution: Express the following percentage as fraction.

1 100 100 1
(i) 11 %= %= =
9 9 9 × 100 9
1 125 125 5
(ii) 62 %= %= =
2 2 2 × 100 8
30 3
(iii) 30% = =
100 10 Rack your Brain
40 2
(iv) 40% = =
100 5 The time duration of 4 hours 45
minutes is what % of a day?
Percentage

3.
 Concept 3: Conversion of a percentage into a ratio: To convert a percentage
into a ratio, first convert the given percentage into a fraction in simplest
form and then to a ratio.

Example 3: Solve the following % values - Previous Year’s Question

(i) 72% (ii) 35%

In a group of people, 28% of the
(iii) 90% (iv) 33.33% members are young while the rest
are old. If 65% of the members are
Solution:
literates, and 25% of the literates
72
(i) 72% = = 18 : 25 are young, then the percentage of
100 old people among the illiterates
35 7 is
(ii) 35% = = ⇒ 7 : 20
100 20 (1) 59% (2) 62%
90 9 (3) 66% (4) 55%
(iii) 90% = = ⇒ 9 : 10
100 10
100
1 100 1
(iv) 33.33% = 33 % = % = 3 = ⇒ 1:3
3 3 100 3

Concept 4: Conversion of a ratio into percentage: To convert a ratio into

percentage, first convert the given ratio into a fraction then to a percentage.

Example 4: Convert the following ratios in percentage:

(i) 2 : 5 (ii) 2 : 3
(iii) 2 : 9 (iv) 3 : 7
Solution:
2
(i) 2 : 5 = × 100 = 40%
5
2
(ii) 2 : 3 = × 100 = 66.66%
3
2
(iii) 2 : 9 = × 100 = 22.22%
9
3
(iv) 3 : 7 = × 100 = 42.84%
7
Percentage

4.
Important Fraction-Percentage Table :

1
1 → 100% → 7.1428%
14
1
→ 50% 1 2
2 → 6 % = 6.66%
15 3
1 1
→ 33 % = 33.33 % 1 1
3 3 → 6 % = 6.25%
16 4
1
→ 25% 1
4 → 5.88%
17
1
→ 20% 1
5 → 5.55%
18
1 2
→ 16 % = 16.67 % 1
6 3 → 5.26%
19
1 2
→ 14 % = 14.28% 1
7 7 → 5%
20
1 1
→ 12 % = 12.5% 1
8 2 → 4.74%
21
1 1
→ 11 % = 11.11% 1
9 9 → 4.5454%
22
1
→ 10% 1
10 → 4.35%
23
1 1
→ 9 % = 9.09% 1
11 11 → 4.166%
24
1 1
→ 8 % = 8.33% 1
12 3 → 4%
25
1
→ 7.69%
13

Rack your Brain

Find 83.33% of 360+45.45% of

660=?
Percentage

5.
 Some Important Related-Fractions:
Rack your Brain
 1
12.5% = 8 
  Find 62% of 132+52% of 440 +
37.5% = 3   2 1
21.25% of 300?
 8 14 7 % = 7 
   
62.5% = 5  57 1 % = 4 
 8  7 7 
 7
87.5% = 
 8

 1 1 
11 9 % = 9 
 
44 4 % = 4 
 9 9 

Concept 5 : Basic tools of percentage : Any number which is independent

of other, that number will always be considered as 100%.
Let us take one number 420.
So, 420 is always considered as 100%
420 = 100%
42 = 10%
4.2 = 1%
0.42 = 0.1%
So these are basic tools, if we want to do calculations on our fingertips.

Example 1: Find 21% of 54.

Solution: Since
54 = 100%
5.4 = 10%
0.54 = 1%
These things should be calculated mentally.
Hence,
Percentage

6.
Example 2: Find 32% of 620.

Solution: Since we know that

Concept 6 : % increase or decrease : Key Note

Case 1 : Both the examples must be solved

mentally by using basic tools of
percentage. We do not need pen
and paper for that.

Absolute value ↑
% Increase = × 100
Original value
Case 2 :

Absolute value ↓
% Decrease = × 100
Original value

Example 1: In a class, 20 girls are there. If 4 new girls joined the class.
Then find % increase in the number of girls. If total students
were 25 in the class initially?

Absolute ↑
Solution: % increase = × 100
Original value

4
% increase = × 100 = 20%
Percentage

20

7.
 Example 2: In a class of 20 girls, 20% girls increase. Then find the total
number of girls after increment.
Solution: New value = Old value + Old value × (% ↑ or ↓)
New value = old value (1 + % ↑ or ↓)
= 20(1 + 20% ↑)
 20 
= 20  1 +  = 20 + 4 = 24
 100 

Alternate Solution:
+20%
20 24
( +4)

Key Note

Note :
We have to use basic tools 20 = 100%
2 = 10%
4 = 20%
Or

Now in the case of decrement :

New value = Old value – Old value × (% ↓)
= Old value (1 – % ↓)
Percentage

8.
Example 3: In a school total number of students are 500. Due to Covid
situation 30% students are not coming to the school. Find
how many students are coming to school currently.
Solution: New value = Old value (1 − % ↓)
= 500 (1 − 30%)
 30 
= 500  1 − 
 100 
= 500 – 150 = 350

Concept 7: Multiplying factor :

In multiplying factor, base always be considered as 1.

Rack your Brain

If the length and breadth of

So first of all we have to know about the multiplying
rectangular plot are increased
factor (What is the multiplying factor?).
by 6.66% and 5.55% respectively.
Let x be a number which decreases by 12.5%. So
Then the new area is how much
we have to find its multiplying factor.
percent more than the original
– 12.5% 1 area?
x x − of x
1 8
↓
8
 1
= x 1 − 
 8
7
= x × → This is multiplying factor .
8

2
Example 1: If a number 3636 is decreased by 16 %. Then find its value
3
after decrement?
2
−16 % = 16.66%
3 5
3636 3636 × = 3030
Solution: 1 6
↓
6
 1 5
Multiplying factor =  1 −  =
 6 6
Percentage

9.
 Example 2: If 770 is increased by 14.28%. Then find its value after
increment, using multiplying factor method?
Solution:
14.28% ↑ 8
770 770 × = 880
1 7
↑
7
↓
 1 8
Multiplying factor 1 +  =
 7  7

Concept 8 : Percentage of a quantity :

Example 1: Find the number of female students, if there are 54% female
students in the school and total number of students in the
school is 800.
Solution: Required number of female students = 54% of 800 = 432

Alternate Method :

Example 2: A student scored 90% marks. If total marks were 450. Find
how much did he score.
Solution: Marks scored = 90% of 450
90
= × 450 = 405
100
1
Example 3: In a garden 33 % of the trees are mango trees. If total
3
number of trees in the garden are 90. Find number of other
types of trees?
Solution: Total number of trees in garden = 90
1
Number of mango trees = 33 % of 90
3
1
= × 90 = 30
3
Percentage

Therefore, the number of other trees = 90–30 = 60

10.
Alternate Method :
→ Mango trees
1 1
33 %=
3 3 → Total number of trees
3 unit → 90
1 unit → 30 (Mango trees)
Therefore, other type of trees = (3–1) unit = 2 unit
= 2 × 30 = 60 trees

Concept 9 : Expressing one quantity as a percentage of another quantity.

Example 1: Raghav obtained 650 marks out of a maximum of 800 marks.

Find the % of marks obtained by him ?
650
Solution: Required percentage = × 100 = 81.25%
800
Example 2: In a company, 600 employees are there. If 40 employees
were absent on a certain day. Then find what % of employees
were present?
Solution: Present employees = 600 – 40 = 560
560
Required % = × 100 = 93.33%
600

Rack your Brain

A rainy day occurs once every 8 days. Half of the rainy days produce
rainbows. What is the percentage of all the days, when we have no
rainbows in a year of 365 Days?

Concept 10 : Advance concept of percentage change :

(a) If a value ‘p’ is increased by x%, then we have to decrease the resultant
 x 
value by  × 100  % to get back to the original value ‘p’.
 x + 100 
Percentage

11.
   100 + x  
P  100  - P 
  
Now the percentage decrease =  × 100
 100+x 
P 
 100 
 x 
= × 100  %
 100 + x 
n
In other words (i.e. in terms of fraction) if a value increased by
then to get
d
back the same number ‘p’ from the resultant value, we have to decrease
 n 
the increased value by  .
 d+ n 

Example 10: Salary of Rahul in 2019 was $200 per day and his salary in
2020 was $ 225 per day. Again in 2021 due to Covid and after
lock down his salary becomes $200 per day.
(i) What is the % increase in his salary in 2020?
(ii) What is the % decrease in his salary in 2021 over 2020?
225 − 200
Solution: (i) % increase in the salary = × 100 = 12.5% ↑
200
(ii) % decrease in the salary in 2021 over
225 − 200
2020 =
225
25 1
= × 100 = × 100 = 11.11%
225 9
In fact, there is same absolute change but percentage
change is different due to different denominators (i.e. base
change). So the base change is as much important as the
numerator or absolute change in quantity.
Example 2: Dheeraj has 33.33% more chocolates than Ankur has. By how
much % less chocolates Ankur has than that of Dheeraj?
Solution:
Percentage

→ So Ankur has 25% less chocolates than Dheeraj.

12.
Example 3: Height of Ayushi is 25% greater than the height of Neeti.
Height of Neeti is how much % less than that of Ayushi?
Solution:
1
+25% =
4

1 1
–20% = =
5 4+1

Thus the height of Neeti is 33.33% less than Ayushi.

(B) If a value ‘p’ is first decreased by x% then to get back the
original value ‘p’ we have to increase the decreased
 x 
(resultant) value by  × 100 
 100 - x 

  x 
P- P  1 - 100  
 
Now the % increase =  × 100
 x 
P1- 
 100 
 x 
= × 100  %
 100 − x 
n
In terms of fraction, if a value ‘p’ is first decreased by then
d
to get back the original number ‘p’. We have to increase the
n
decreased (or resultant) value by .
d- n
Example 4: Due to irregular working habits of Rajesh, his salary was
reduced by 20% but after sometime his salary was increased
to the original salary. What is the percentage increase in
Percentage

salary of Rajesh?

13.
 Solution: Let initial salary be Rs. 100.
Rack your Brain

In a village, the production of

food grains increased by 40%
and per capita production of food
grains increased by 27% during a
certain period. The percentage
by which the population of the
Alternate Solution: village increased during the same
period is nearest to ?

20
% increase = × 100 = 25% ↑
80

Key Note

1. If a value ‘x’ is first increased by p% to ‘y’ then ‘y’ is again decreased
to ‘x’ by q%. Then ‘p’ is always greater than ‘q’ (for positive values).
2. If a value ‘x’ is first decreased by p% to ‘y’ and then ‘y’ is increased
by q% to ‘x’, then ‘p’ is always less than ‘q’.
3. If a value ‘A’ is increased by p% then again by q% once again it is
increased by r%, then the final value will be same as if you change
the order of p, q, r i.e. A can be first increased by r% and then by
q% and then by p% still the result will be same.
4. Rule 3 is also applicable for the decreasing of the values. A value ‘A’
is first decreased by p% then by q% and then by r% and so on. The
resultant value will be same as when ‘A’ is first decreased by q%.
Then by p% and then by r% etc.

Note: In case 3 and 4 we are discussing the successive increase or

decrease in the value.

5. A value ‘A’ is first increased by p% then by q% and then it is reduced

by r% will give the same results as when ‘A’ is first decreased by r%.
Then increased by q% and then by p% etc.
Percentage

14.
Example 1: Initially Katrina has Rs. 600. Then she increased it by 40%.
Once again she increased her amount by 25%. Then final
value of money she has in her wallet will be how much %
greater than the initial amount ?
Solution:

Example 2: The salary of Anil is 50% greater than the salary of Sunil. The
salary of Ankit is 20% less than the salary of Anil. By how
much% the salary of Ankit is greater than the salary of Sunil?
Solution:

20
So the required percentage change = × 100 = 20%
100

Concept 11: Reversing the change by same % : If the value of a number is

first increased by x% and then decreased by x%. The net change is always
a decrease (or loss)in original value.
2
 x  x2
That is, % loss =   % = %
 10  100

Explanation: Let original number be 100.

100 x x2
= 100 + x− −
100 100
x2
= 100 + x− x−
100
Percentage

15.
 Previous Year’s Questions

In 2010, a library contained a total

of 11500 books in two categories
- fictional and non-fictional.
In 2015, the library contained a
total of 12760 books in these two
categories.
During this period, there was 10%
increase in the fiction category
while there was 12% increase in
the non-fiction category. How
many fiction books were in the
library in 2015?
(1) 6600 (2) 6160
(3) 6000 (4) 5500

Example 1: Saurav is an expert in bargaining. Once he went to the nearby

shop, when Saurav asked the price of a shirt the shopkeeper
told him the price by increasing 20% of the original cost.
But Saurav insisted to decrease the price by 20%. So the
shopkeeper sold it by decreasing the price by 20%. What is
the loss or profit of shopkeeper and by how much % ?
(1) No loss
(2) No profit
(3) Profit of 1.5%
(4) Loss of 4%
Solution: Let the actual price be 100.
Percentage

16.
Alternate Solution:
2
 20  400
Loss % =   = = 4%
 10  100
There is always a loss.

Example 2: If the length and breadth of a rectangle are changed +25%

and –10%, what is the % change in area of rectangle ?
(1) 12.5% (2) 37.5% (3) 50% (4) 25%
Solution: Area of rectangle = Length × Breadth
+1 −1
25% = 10% =
4 10
Previous Year’s Questions

In a class, 60% of the students

are girls and the rest are boys.
There are 30 more girls than boys.
If 68% of the students, including
30 boys, pass an examination,
5 the percentage of the girls who
% increase in area = × 100 = 12.5% ↑
40 do not pass it?  [TITA]
Alternate Solution:
l×b = area
Let length is 1 unit and breadth is 1 unit.
l ×b = Area
Percentage

So there is 12.5% increase in the area of rectangle.

17.
 Example 3: P’s annual income is 20% lower than Q’s annual income and
R’s annual income is 56.25% greater than P’s annual income.
By how much % Q’s annual income is less than R’s annual
income?
(1) 20% (2) 40% (3) 33.33% (4) 9.09%

Solution: Let the annual income of Q be 100.

25
The required value = × 100 = 20%
125

Concept 12 : Product constancy : It is same as the inverse proportion. For

example when the price of rice is Rs. 40/kg we can purchase 10 kg of rice.
But when the price of rice increases and it become Rs. 50/kg then we can
purchase only 8 kg of rice. So if we check our expenditure we will get same
expenses as the previous one.

So the product of price and consumption is constant in both the cases.

Percentage

18.
 Some more examples of product constancy :
(i) Speed × Time = Distance
(ii) Price × Consumption = Expenditure
(iii) Efficiency × Time= Constant
(iv) Length × Breadth = Area
(v) Rate × Time = Constant
(vi) Average × number of observation = Sum of all observation

Example 1: The price of sugar is increased by 20%. Then by how much

% should customer reduce the consumption (Quantity used)
of sugar so that he does not need to increase his expenses
on sugar?
Solution:

Rack your Brain

A family decreases its

consumption of pulses by 42.84%.
Because the price of pulses has
10000 500 250 1 increased by K%. Find K?
⇒x= = = = 83 kg
120 6 3 3 (If expenditure remains constant)
1
Therefore, % reduction = 100 − 83 = 16.67%
3
Or [P×C = E]
1×1=1
1.2 × k = 1
1 5
k= =
1.2 6
5 1 2
Hence, % reduction = 1 − = = 16 %
6 6 3
Thus, there will be 16.67% decrease in the consumption of sugar in
order to maintain the same expenditure on sugar.

Product Constancy Conditions:

yy When one factor of a product is increased by p% then the other factor will
 P 
be decreased by  × 100  %
 100 + P 
n
It means when one factor of a product is increased by then the other
Percentage

d
n
factor is decreased by .
(d+ n)
19.
 yy When one factor of a product is decreased by P% then the other factor will
 P 
be increased by  × 100  %
 100 - P 
n
It means when one factor of a product is decreased by . Then the other
d
 n 
factor must be increased by  
 d- n 
Example 2: If the price of petrol is increased by 25%. By how much %
must a person decrease his consumption so that expenditure
remains the same?
Solution: Since product is constant

Increases by Decreases by
1 1
25% = ↑ ↓ = 20%
4 5

1 1
If anything is increased by ↑ then it must be decreased by ↓ and
n n+ 1
vice versa.

Alternate Solution:
P × C = Expenditure (Since expenditure will remain same)
1
P∝
C
+1
25% =
4

1
% reduction in consumption = × 100 = 20%
5
Example 3: Due to 50% increase in the price of rice. We can purchase
5 kg less rice with the same amount of Rs. 100. What is the
new price of rice ?
(1) Rs. 20/kg (2) Rs. 10/kg
(3) Rs. 50/kg (4) Rs. 60/kg
Solution: P×C = E
1 1
P∝ 50% =
Percentage

C 2

20.
 Rs . 100 20
Hence, old price of rice = = = Rs . 6.66 / kg
15 3
Rs . 100
New price of rice = = Rs. 10 /kg.
10

Concept 13 : Difference between “by” and “To” : There is a clear difference

between “by” and “to”. For example the income of Raju is increased by 40%
which means the new income is 140% of the original income and if we say
that income is reduced to 40% would mean the new income is 40% of the
original income.
Example 1: If the income of Priya is increased by 30%. It means her new
income is 100 + 30 = 130% of the original income. If we say
that income of Priya has increased to 30%. It means the new
income of Priya is 120% of the original income.
Example 2: Two numbers are respectively 25% and 40% less than a 3rd
number. What % is the 2nd of the first ?
Solution: Let 3 numbers be A, B and C and 3rd number be 100.
A : B : C
75 : 60 : 100
60
Required % = × 100 = 80%
75
Concept 14: Population based : If the original population of a locality be ‘P’
and the annual growth rate be r%.
n
 r 
The population after ‘n’ years = P  1 + 
 100 
n
 r  
Increase (change) in the population = P  1 +  − 1
 100  
If there is decrease in population by r% then total population after ‘n’ years
n
 r 
= P1- 
 100 
n
  r  
Percentage

And decrease in population = P 1 -  1 -  

  100  

21.
 Example 1: If the present population of a city is 50,000 and there is 20%
increase in the population every year. Then what will be the
population after 3 years ?
(1) 86400 (2) 76400
(3) 86700 (4) 90000
Solution: Population after 3 years
 r 
n
Rack your Brain
= P1 + 
 100 
20 
3
The present population of a
= 50000  1 +  country is 10 crores. If it rises to
 100 
17.28 crores during next 3 years,
6 6 6
= 50000 × × × then find the uniform rate of
5 5 5
growth in population?
6 6 6
= 50000 × × ×
5 5 5
= 86400
Hence, option (1) is the correct answer.

Previous Year’s Question

In an examination, the maximum possible score is N while the pass

mark is 45% of N. A candidate obtains 36 marks, but falls short of the
pass mark by 68%. Which one of the following is then correct?
(1) N ≤ 200 (2) 243 ≤ N ≤ 252
(3) N ≥ 253 (4) 201 ≤ N ≤ 242

Example 2: The population of a town in the first year increases by 10%. In

the next year it decreases by 10%. Once again in the 3rd year
it increases by 10% and in fourth year it decreases by 10%. If
present population is 20,000, then find the population after
four years?
(1) 19805 (2) 18502 (3) 19602 (4) 21500
Solution: Population after 4 year
th

 10   10   10   10 
= 20000  1 +  1 −  1 +  1 − 
 100  100  100  100 
Percentage

22.
 11 9 11 9
= 20000 × × × ×
10 10 10 10
= 20000 × 1.1 × 0.9 × 1.1 × 0.9 = 19602
Thus option (3) is the correct answer.

Rack your Brain

The pollution in normal atmosphere is less than 0.01%. Due to leakage

of methyl isocyanate (MIC) from a factory, the pollution is increased to
20%. If every day 80% of the pollution in the atmosphere is neutralized,
in how many days will the atmosphere become normal ?

Percentage

23.
 Practice Exercise

Easy :

1. 
A wire manufacturing company purchased a certain amount of raw mate-
rial, of which 10% was wasted due to improper handling. After using 85%
of the remaining raw material, 47.25 kg of raw material was left. How much
raw material was purchased initially?
(1) 200 kg (2) 250 kg (3) 300 kg (4) 350 kg

2. 
Rate of inflation is 2000% per annum. What is the value of the article two
years from now if it costs Rs. 12 today?
(1) Rs. 5292 (2) Rs. 5992 (3) Rs. 6992 (4) Rs. 7052

3. 
3 friends A, B and C donate 8%, 7% and 9% of their salary to a charitable
hospital in the given order. Salary of A and B is same and the difference of
their donation is Rs. 74. The total donation by A and B is Rs. 525 more than
‘C‘s donation. What is the percentage of the total salary of the 3 donations?
(1) 7.85% (2) 7.95% (3) 6.34% (4) None of these

4. 
Due to a price hike of 9.09%, 2 kg less sugar is available for Rs. 60. What is
the initial price per kg of sugar?
(1) Rs. 5/kg (2) Rs. 2.5/kg (3) Rs. 10/kg (4) Rs. 9/kg

5. 
Kajal went to a vegetable market with a certain amount of money, with this
money, she can buy either 40 apples or 70 mangoes. She retains 15% of her
money for auto fare. If she buys 35 mangoes, how many more apples can she
buy?
(1) 14 (2) 20 (3) 25 (4) 28

Moderate:

6. 
A’s income is 75% of B’s income, and A’s expenditure is 80% of B’s expend-
iture. If A’s income is 90% of B’s expenditure, then find the ratio of A’s sav-
ings to B’s savings.
(1) 1 : 2 (2) 2 : 1 (3) 5 : 2 (4) 3 : 5

7. 
The annual earning of Mr. Ashwani Lal is Rs. 4 lakhs per annum for the first
year of his job and his expenditure was 50%. Later on for the next 3 years
his average income increases by Rs. 40,000 per annum and the saving
was 40%, 30% and 20% of the income. What is the percentage of his total
Practice Exercise

savings over the total expenditure if there is no interest is applied on the

savings for these four years?
(1) 49 37 % (2) 41 73 % (3) 53% (4) 51 6 %
87 83 19

24.
8. 
A man deposited 14% of the initial amount to his locker and again after
some time he deposited 45% of the increased amount becomes Rs. 16530.
How much was the initial amount?
(1) Rs. 12000 (2) Rs. 16000 (3) Rs. 18000 (4) Rs. 10000

9. 
Suraj gave 25% of his monthly salary to his mother from remaining salary
he paid 15% towards rent and 25% he kept for his monthly expenses. The
remaining amount he kept in a bank account. The sum of the amount he
kept in bank and that he gave his mother was Rs. 42000. What was his
monthly salary?
(1) Rs. 48000 (2) Rs. 62000 (3) Rs. 60000 (4) Rs. 50000

10. The income of Mr. Rajesh is 88.88% more than his expenditure. If his in-
come increases by 30% and his expenditure increases by 35%. Then his
saving increases by Rs. 78000. Find the difference between his initial in-
come and initial expenditure?
(1) Rs. 320000 (2) Rs. 440000 (3) Rs. 550000 (4) Rs. 660000

Difficult:

11. 
There are two types of biscuits P and Q, type P biscuit has carbohydrates,
fats and proteins and type Q has carbohydrate, fats and sugar. These two
types of biscuits are mixed to form type ‘R’ biscuit. Type ‘P’ and ‘R’ has 30%
carbohydrates. The % of sugar in type ‘Q’ and ‘R’ is 40% and 20% respec-
tively. Type ‘R’ has 24% proteins. The % of fat in type ‘R’ biscuit is ?
(1) 26% (2) 33% (3) 21% (4) 22%

12. A sweet seller decreases the price of 1 kg sweet by 20% in 1st month, in
2nd month he increases the price of the 1kg sweet by 20%. He follows this
particular pattern in 3 and 4, 5 and 6 month and so on. Find the month in
which the price of the sweet per kg goes below 60% of the initial price for
the first time.
(1) 17th month (2) 16th month (3) 8th month (4) 10th month

13. In a class, each student likes at least one of the five subjects namely Phys-
ics, Chemistry, Maths, English and Biotechnology. 80% of the students like
Physics subject, 70% of the students like Chemistry, 90% of the students
like Maths, 80% of the students like English subject and 90% like Biotech-
Practice Exercise

nology. What is the maximum percentage of the students who like exactly
4 of the 5 subjects?
(a) 90% (b) 70% (c) 62% (d) 50%

25.
 14. In every month Ravindra consumes 25 kg rice and 9 kg wheat. The price
of rice is 20% of the price of wheat and thus he spends total Rs. 350 on
the rice and wheat per month. If the price of wheat is increased by 20%
then what is the percentage reduction of rice consumption for the same
expenditure of Rs. 350? Given that the price of rice and consumption of
wheat is constant.
(1) 36% (2) 40% (3) 25% (4) 24%

15. In a factory there are three types of Machines M1, M2 and M3 which produces
25%, 35% and 40% of the total products respectively. M1, M2 and M3 produc-
es 2%, 4% and 5% defective products, respectively. What is the percentage
of non-defective products?
(1) 89% (2) 97.1% (3) 96.1% (4) 86.1%

ANSWER KEY

1. (4) 2. (1) 3. (2) 4. (2) 5. (1)

6. (1) 7. (4) 8. (4) 9. (3) 10. (1)

11. (1) 12. (1) 13. (1) 14. (1) 15. (3)

SOLUTIONS

Sol. 1 (4); 350 kg

Let 100x of raw material is purchased initially.

10% of raw material was wasted due to improper handling so that remaining
raw amount of material after wastage = 100x – 100x × 10% = 90x
Since, 85% of the remaining raw material is used for the production.
Hence, remaining amount of raw material = 90x – 90x × 85% = 13.5x
→ 13.5x = 47.25 kg
47.25
x= = 3.5kg
13.5
∴ Amount of raw material purchased initially was = 100x = 100 × 3.5 = 350 kg

Alternate Solution:
Practice Exercise

Let amount of raw material purchased initially was = 100 unit

26.
13.5 unit → 47.25kg
47.25
1unit → = 3.5kg
13.5
∴ 100unit → 100×3.5 = 350 kg
Hence, option (4) is the correct answer.

Sol. 2 (1); Rs. 5292

Since this question can be done by various methods like straight line
method, ratio method or compound interest method.
If we solve it by straight line method.
12 →
+2000%
12 + 12 × 2000%
= 12 + 240 = 252 (At the endof first year)

252 →
+2000%
252 + 252 × 2000%
= 252 + 5040
= 5292 (At the endof second year)
Hence, option (1) is the correct answer.
Alternate Solution:
Practice Exercise

27.
 Sol. 3 (2); 7.95%

Let salary of A and B is Rs. x each and salary of ‘C’ is Rs. y.

(8x% – 7x%) → Rs. 74

x% → Rs. 74
x
→ Rs. 74
100
x → Rs. 7400

Hence, donation of ‘C’ = Rs. 1110 – 525 = Rs. 585

Since, donation of ‘C’ is 9y%
Then, 9y % → 585

y 585
→
100 9
y → Rs. 6500
Now we have to find total donations of A, B and C together

= 592 + 518 + 585 = Rs. 1695

Total salary of A + B + C = 7400 + 7400 + 6500 = Rs. 21300

1695
Required % = × 100 = 7.95%
21300
Practice Exercise

Sol. 4 (2); Rs. 2.5/kg

+1
9.09% =
11

28.
 60
∴ Initial price of the sugar = = Rs . 2.5 / kg
24

Sol. 5 (1); 14

Let she had = Rs. 100

Cost of 40 apples = cost of 70 mangoes = Rs. 100

70 mangoes = Rs. 100
35 mangoes = Rs. 50
Remaining money = Rs. 85 – 50 = Rs. 35
Since Rs. 100 → 40 Apples
40
Now for Rs. 35 → × 35 Apples
100
2
= × 35 = 14 Apples can be bought
5
Hence, option (1) is the correct answer.

Sol. 6 (1); 1 : 2

Let B’s income is Rs. x

A’s income = 75% of B’s income
3 3
A’s income = × x= x
4 4
Also let B’s expenditure = Rs. y
4
Then A’s expenditure = 80% of B’s expenditure = ×y
5
Practice Exercise

Since, A’s income = 90% of B’s expenditure

29.
 3 9
x= ×y
4 10
x 9×4 6
= =
y 3 × 10 5
6
x= y
5
Now we have to find ratio of their savings.

3 4
x- y
A's savings 4 5
=
B's savings x- y
3 6 4
× y- y
= 4 5 5
6
y- y
5

18y 4 2y
− y
10 1
  = 20 5 = 20 = =
6y − 5y y 20 2
5 5
Hence, ratio of their savings is 1 : 2
Option (1) is the correct answer.

6
Sol. 7 (4); 51 %
19
Income → [4 4.4 4.8 5.2] → 18.4 Lakh
Saving → [2 1.76 1.44 1.04] → 6.24 Lakh
Exp. → [2 2.64 3.36 4.16] → 12.16 Lakh

6.24 6
So, × 100 = 51 %
12.16 19
Practice Exercise

30.
Sol. 8 (4); Rs. 10000

Let initial amount be 100 unit.

100 Unit
+14%

114 Unit Increased amount

+45% [51.3 Unit] 114 = 100%
165.3 Unit 11.4 = 10%
45.6 = 40%
5.7 = 5%
51.3 = 45%

Now, 165.3 Unit Rs. 16530

1 Unit 165300 = Rs. 100
16503

∴ 100 unit → 100 × 100 = Rs. 10000

Hence, initial amount is Rs. 10000.

Sol. 9 (3); Rs. 60000

Let monthly salary of Suraj is Rs. x

3 3 1
x× × + x× = Rs . 42000
4 5 4
9x x
+ = Rs . 42000
20 4
14 x
= 42000
20
x = Rs . 60000

Alternate Solution:
Practice Exercise

31.
 LCM of (4, 3, 20) = 60
Let total salary of Suraj is 600 unit.

Gave to Mother + Bank Account

(150 + 270) unit → Rs. 42000
420 unit → Rs. 42000
1 unit → Rs. 100
600 unit → 100 × 600 = Rs. 60000

Sol. 10 (1); Rs. 320000

We know that :

195 Unit → Rs. 78000

78000
1 Unit → = Rs . 400
195
Practice Exercise

32.
Hence, option (1) is the correct answer.

Sol. 11 (1); 26%

Carbohydrate Fats Proteins Sugar

Biscuit P 30% 22% 48% ×
Biscuit Q x% → 30% 30% × 40%
Biscuit R 30% 26% 24% 20%

Type R biscuit has carbohydrates = Avg. of carbohydrates of biscuit P and Q

30% + x
30% =
2
60% = 30% + x
x = 30%
% of proteins + % of proteins in
in biscuit P in biscuit Q
% of proteins present in biscuit R =
2
P+0
24% =
2
% of proteins in biscuit P = 48%
Therefore, fat in biscuit “p” = 100%-(30%+48%)
= 22%
Also, % of fat in biscuit Q = 100% – (30% + 40%)
= 30%
Hence % of fat present in biscuit R is average of % fat present in biscuit P
and % of fat present in biscuit Q, because biscuit ‘P’ and ‘Q’ are mixed in
(1 : 1).
22% + 30%
∴ % of fat in biscuit R = = 26%
2
Hence, option (1) is the correct answer.
Practice Exercise

33.
 Sol. 12 (1); 17th month

Let the price of 1kg sweet initially = ₹ 100.

After decrement of 20% in 1st month the price becomes = 100 × 80% = ₹
80/kg.
In 2nd month it is increases by 20% again = 80 × 120%
= ₹ 96/kg
In 3rd month it is again going to be decreased by 20% = 96 × 80% = ₹ 76.8/
kg
Again in 4th month the price per kg of sweet is going to increase by
20%.
= 76.8 × 120%
= ₹ 92.16/kg
We can observe here:

3.84
Overall % decrement in the price of sweet = × 100
96
= 4% ↓
After 4 months calculation, what we can observe, easily that after every 2
month the price of the 1 kg sweet decreases by 4%.
⇒ (0.96)n
At n = 8, the value is 0.7211 and in next month the price of the sweet goes
down by 20% ⇒ Less than 60% of the initial price.
Hence, 8 × 2 +1 = 17th month
Or we can say that after 1 year 5 months, The price of 1 kg sweet will be less
than 60% of the initial price.
Practice Exercise

Hence option 1 is the correct answer.

34.
Sol. 13 (1); 90%

Let ‘M’ be the numbers of the students who like 1 subject.

‘N’ be the numbers of the students who like 2 subjects.
‘O’ be the numbers of the students who like 3 subjects.
‘P’ be the numbers of the students who like 4 subjects.
‘Q’ be the numbers of the students who like 5 subject.
Also, We know that:

Therefore M × 1 + 2 × N + 3 × O + 4 × P + 5 × Q = 410%

Now, We have to find the maximum % of the students who like exactly 4 of
the 5 subjects.
Therefore, let ‘M’ = 0
‘N’ = 0
‘O’ = 0
In ideal condition ‘Q’ should also be considered as ‘O’ but if you consider Q
= O, then 2nd equation will never satisfy the condition.
Hence, Q ≠ 0 must be considered.
Practice Exercise

∴ P + Q = 100% …(1)
4P + 5Q = 410% …(2)

35.
 Now solve equation (1) and (2) we get:
4P + 4Q = 400%
4P + 5Q = 410%
–Q = –10%
Q = 10%

Then P + Q = 100%
P + 10% = 100%
P = 90%
Hence the correct answer is option (a).

Sol. 14 (1); 36%

Rice Wheat
25 9
↓ ×x × 5x
25 x 45 x
70x = 350 ⇒ x = 5
Hence, the price of Rice = Rs. 5/kg.
Price of wheat = Rs. 25/kg.
Now, the price of wheat = Rs. 30/kg.
Let the new amount of Rice be M kg.,
Then, M × 5 + 9 × 30 = 350 ⇒ M = 16
25 − 16
Hence, decrease (in %) of amount of rice = × 100 = 36%
25
Sol. 15 (3); 96.1%

Non-Defective products

25 × 0.98 + 35 × 0.96 + 40 × 0.95

× 100 = 96.1 %
100
Practice Exercise

36.
MIND MAP

MIND MAP

37.